BOD and COD Calculations in Environmental Chemistry
Biochemical Oxygen Demand (BOD) and Chemical Oxygen Demand (COD) are two fundamental water-quality parameters tested in environmental chemistry for Class 11, Class 12, JEE Main, NEET, and CUET. In aquatic ecosystems, aerobic organisms rely on dissolved oxygen (DO) to survive. When organic pollutants enter water reservoirs, aerobic microorganisms consume dissolved oxygen to decompose the waste. Measuring BOD and COD enables chemists to determine pollution levels and assess wastewater treatment efficiency.
Understanding BOD and COD: Principles and Chemistry
Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen consumed by aerobic microorganisms to biologically degrade organic matter in a water sample over a standard incubation period, typically 5 days at 20 °C (termed BOD5). Clean drinking water has a BOD below 1 mg/L, and clean natural water remains below 5 mg/L (ppm). In contrast, polluted water exhibits BOD values exceeding 17 mg/L, with untreated domestic sewage often reaching several hundred mg/L.
Chemical Oxygen Demand (COD) measures the total quantity of oxygen required to chemically oxidize all oxidizable compounds in a water sample. Unlike BOD, COD measures both biodegradable and non-biodegradable organic substances. In standard laboratory analysis, the water sample is refluxed with a known excess of potassium dichromate (K2Cr2O7) in concentrated sulfuric acid, using silver sulfate (Ag2SO4) as a catalyst and mercuric sulfate (HgSO4) to eliminate chloride interference. The unreacted dichromate is back-titrated with standard ferrous ammonium sulfate (FAS, Mohr's salt) using ferroin indicator.
Because strong chemical oxidation degrades resistant compounds that bacteria cannot break down, the COD of any sample is always greater than or equal to its BOD (COD ≥ BOD).
Key Formulas and Calculation Methods
BOD calculations involve dissolved oxygen depletion adjusted for dilution, whereas COD calculations rely on redox titration stoichiometry.
BOD and COD Working Formulas:
Unseeded BOD5: BOD5 (mg/L) = (DO_i - DO_f) × DF = (DO_i - DO_f) / P
Seeded BOD5: BOD5 (mg/L) = [(DO_i - DO_f) - (B_i - B_f) × f] / P
COD: COD (mg/L) = [(V_blank - V_sample) × M_FAS × 8 × 1000] / V_sample
Where:
- DO_i = Initial dissolved oxygen of diluted sample in mg/L
- DO_f = Final dissolved oxygen of diluted sample after 5 days in mg/L
- P = Dilution fraction = (Wastewater volume) / (Total bottle volume)
- DF = Dilution factor = 1 / P = (Total bottle volume) / (Wastewater volume)
- B_i = Initial dissolved oxygen of seed blank in mg/L
- B_f = Final dissolved oxygen of seed blank after 5 days in mg/L
- f = Seed volume fraction = (Total volume - Wastewater volume) / Total volume
- V_blank = Volume of FAS used for blank titration in mL
- V_sample = Volume of FAS used for wastewater titration in mL
- M_FAS = Molarity of ferrous ammonium sulfate in mol/L
- 8 = Equivalent mass of oxygen in g/equiv (32 g/mol divided by 4 electrons transferred per O2)
- 1000 = Conversion factor from grams to milligrams
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, CBSE Class 12, JEE Main, NEET, CUET
- Topic: BOD and COD Calculations in Environmental Chemistry
- Difficulty: School to foundation competitive level
Step-by-Step Solved Examples
The following solved numerical problems illustrate standard exam questions. Every arithmetic step is displayed in full, followed by verification using an alternative method.
Question: A 15.0 mL sample of municipal wastewater is diluted to 300.0 mL in a BOD incubation bottle using aerated distilled water. The initial dissolved oxygen of the mixture is 8.60 mg/L. After incubation at 20 °C for 5 days, the dissolved oxygen is 3.20 mg/L. Calculate the 5-day biochemical oxygen demand (BOD5) of the wastewater sample in mg/L.
Step 1: Compute dissolved oxygen depletion in the bottle. DO depletion = DO_i - DO_f = 8.60 - 3.20 = 5.40 mg/L.
Step 2: Determine the dilution factor (DF). DF = Total volume / Sample volume = 300.0 / 15.0 = 20.0.
Step 3: Calculate BOD5 of the wastewater. BOD5 = DO depletion × DF = 5.40 × 20.0 = 108.0 mg/L.
Verification via alternative route: Dilution fraction P = 15.0 / 300.0 = 0.0500. BOD5 = (DO_i - DO_f) / P = 5.40 / 0.0500 = 108.0 mg/L.
Answer: 108.0 mg/L
Question: A 50.0 mL sample of treated effluent is refluxed with 25.0 mL of potassium dichromate solution in acid medium. The unreacted dichromate requires 14.20 mL of 0.100 M ferrous ammonium sulfate (FAS) to reach the endpoint. A blank containing 50.0 mL of pure water requires 24.00 mL of the same FAS solution. Calculate the Chemical Oxygen Demand (COD) of the effluent in mg/L.
Step 1: Calculate titrant difference. ΔV = V_blank - V_sample = 24.00 - 14.20 = 9.80 mL.
Step 2: Milliequivalents of FAS. For FAS (Fe2+ → Fe3+ + e-), n = 1, so Normality = 0.100 N. Milliequivalents = 9.80 × 0.100 = 0.980 meq.
Step 3: Mass of oxygen consumed. Equivalent weight of O = 32 / 4 = 8 g/equiv = 8 mg/meq. Mass of O = 0.980 × 8 = 7.84 mg.
Step 4: Compute COD in mg/L. Sample volume in litres = 50.0 / 1000 = 0.0500 L. COD = 7.84 / 0.0500 = 156.8 mg/L.
Verification via direct formula route: COD = [(24.00 - 14.20) × 0.100 × 8 × 1000] / 50.0 = [9.80 × 800] / 50.0 = 7840 / 50.0 = 156.8 mg/L.
Answer: 156.8 mg/L
Common Mistakes in Water Analysis Calculations
Frequent Student Pitfalls:
- Assuming BOD can exceed COD: Chemical oxidation is more comprehensive than microbial decomposition. Therefore, COD is always greater than or equal to BOD. If a calculation gives BOD > COD for the same water sample, recheck your numbers.
- Omitting dilution factors: Reporting the DO depletion inside the incubation bottle rather than scaling by the dilution factor is a common error. The bottle DO drop reflects only the diluted mixture.
- Using incorrect equivalent mass for oxygen: In COD calculations, students sometimes multiply by 16 or 32 instead of 8. Because oxygen gas (O2, 32 g/mol) accepts 4 electrons (O2 + 4H+ + 4e- → 2H2O), the equivalent weight of oxygen is 32 / 4 = 8 g/equiv.
- Subtracting sample from blank in reverse: The blank consumes more FAS titrant than the sample because none of its dichromate was reduced by organic waste. The volume difference is always (V_blank - V_sample).
Exam Relevance and Question Patterns
| Exam | Question Type | Key Focus Areas |
|---|---|---|
| CBSE Class 11 & 12 | Short answer and definition questions | Differences between BOD and COD, clean water limits (< 5 ppm), polluted water thresholds |
| JEE Main | Single correct multiple choice questions | Unseeded BOD dilution arithmetic, COD redox titration calculations, and pollution status determination |
| NEET | Assertion-Reason and direct MCQs | Ecological significance of dissolved oxygen, BOD comparison between clean and polluted water bodies |
| CUET (UG) | Objective numerical questions | Dilution factor application, standard unit conversions (mg/L to ppm), and COD chemical reagent roles |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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